5 s the line segment from 1 0 to 0 1 can be

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Unformatted text preview: ) S The line segment from (1 0) to (0 1) can be parameterized by 0 ≤ 1 − . On this line segment, part of the integral becomes 1 The line segment from (0 1) to (−1 0) can be parameterized by −1 ≤ ≤ 0 − . On this line segment, part of the integral becomes −1 −1 = −1+ = = −2 (−1 + ) + (− ) =0 |−1+ |+|− | The line segment from (0 −1) to (1 0) can be parameterized by −1 ≤ On this line segment, part of the integral becomes 0 = −1 The line segment from (−1 0) to (0 −1) can be parameterized by 0 ≤ ≤ 1 − . On this line segment, part of the integral becomes 0 = −1− 0 (− 1 − ) + (− ) = −2 |−1− |+|− | 1 = (1 − ) + =0 |1 − | + | | 0 0 ≤1 (1 + ) + =2 |1 + | + | | ≤0 = 1+ . 0 =2 −1 Thus, we conclude that C E + =0−2+0+2=0 | |+| | :) Evaluate 2 2 + + 2...
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